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1 like and no replies? Let me fix that

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To check that a given function y(t) is a solution to the differential equation you want to check that it does in fact do what the DE says, in this case the left hand side says take the derivative of y(t) and the right hand side says to take y(t) and divide it by t^2. Then y(t) is a solution if the left hand side is equal to the right hand side. I suppose you could also say, is the anti-derivative of y(t)/t^2 equal to y(t), but generally that's a more difficult problem and there isn't a unique antiderivative so it's better to go the other way.

@@dr.clarkteachesmath thanks prof. I got it

I think the best answer is that it works and you can show in a more theory driven advanced course why it works, but in a first course, it's more about using the tool to solve DEs than proving why that tool is valid.

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Yes, that's right, it should have kt/3 like you said.

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This sounds like a standard sinking fund problem. The formula for the future value of a sinking fund, that is a regular sequence of deposits at a fixed interest rate is FV = P((1+i)^n - 1)/i where i is the annual interest rate and n is the number of years. In your case, 5000(1.05^10 - 1)/.05 = 62889.5. Since the debt is 60,000 at year 10, that leaves a surplus of 2889.50 at year 10.

@@dr.clarkteachesmath why it is 1.05^10-1?

@@IamAspirant001 That comes from the sinking fund formula. It's a standard formula in financial mathematics. I'd suggest searching for a video that specifically derives the formula and explains where it comes from. The short answer is that it is comes form the formula for summing a series, but I can't fit the explanation in a comment.

The videos are intended to be a very brief introduction to the topic, not a complete explanation. So yes, you are correct that it's not a sufficient explanation, but I didn't intend it to be, only a preview of what is coming up in my FM class.

@@dr.clarkteachesmath Okay. No worries. I found the answers and understanding somewhere, eventually.

There are two versions of Green's Theorem in 2D, one gives you the net circulation around the curve, and that one has the curl F inside the region, and then there is also a flux version which gives the net flux through the boundary to be the sum of the divergence within the curve.

The valued 3/2 is a scaler. It's a measure of the "flux" out the surface of the vector field. So, if the surface is a net, and the vector field is the flow of water, the flux would be how much water is flowing through the net. So the units depend on the units of the vector field - it's could be an electromagnetic field and then you'd have to ask a physicist what the units mean in that case. If you're looking for a nice book that explains divergence and curl in a conceptual way I'd suggest "Div, Grad, Curl, and All That". It's a friendly and short read.

Glad to be helpful.

Yes, that is correct. Good catch!

in cylindrical coordinates, you need to have dV the volume differential, which is r dz dr dtheta. In rectangular coordinates, it is just dV = dx dy dz, a box, but in polar coordinates dx dy = r dr dtheta, so in cylindrical, dV = r dz dr dtheta. You'll need to look at a picture of the polar "rectangle" r dr dtheta from a textbook to see why the r is there.

you're fucking life saver

​@@dr.clarkteachesmath i've got question so if dx,dy,dz = r, dr, dz, because dx, dy = r dr so why do you add "dθ" with the r dr dz dθ 6:42

I don't have a plan to do a PDE series, but I should add more ODE content tbh.

Charles, good point, the video might have been more helpful if a wider variety of shapes were shown. What were you hoping for?

Gunjan, interesting question. I would suggest using the divergence theorem to convert the surface integral for your vector field F into an integral of it's divergence (in this case div F = 2x + 2y + 2z). Since div F is positive and negative exactly half the time symmetrically on the given domain (the solid inside the surface) the integral will come to 0 when computed.

Thanks for watching Kardelen. I'm glad they were helpful.

Thanks for watching Alexis. Glad it was helpful.

Thanks Fezeka. Glad it was helpful.

Andy, sorry for the delayed reply. I do not, but you're welcome to use what I have available here.

Glad it was helpful!

Damn...

You should be able to use it along a particular path, but you would have to check it for all possible paths.

I'm using CalcPlot 3d, which is available free here: www.monroecc.edu/faculty/paulseeburger/calcnsf/CalcPlot3D/ but I also use GeoGebra regularly which is free here: www.geogebra.org/ Both can run in a browser or can be downloaded.